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In higher dimensions I want to find the angles between objects of various dimensions. For example, in 4D it seems to me that if there are two intersecting planes then there have to be two angles between the planes. Or go to 6D with two intersecting 3D hyperplanes. Or how about an intersection of a 4D hyperplane and 2D plane? It seems to me that this should all have been figured out a century ago. Can anyone direct to a reference?

- PatrickPowers
- Tetronian
**Posts:**478**Joined:**Wed Dec 02, 2015 1:36 am

Angles between hyperplanes is actually one of those things that seem totally intuitive, but mathematically it turns out to be a lot more complex than one might expect based on 3D experiences. In 3D, there's always a well-defined minimal angle between two planes, regardless of orientation. In higher dimensions, this may or may not be true. I think even in 4D this may still hold, but once you go higher, say in 7D, then a pair of 2D planes may have multiple possible angles depending on how you measure it. I've seen a couple of papers that try to define various angles between two such objects (in more general terms, it's the angle between two subspaces of an Euclidean space), or to unify the diverse measurements into a single figure, but AFAIK past 4D or 5D or somewhere thereabouts there is no universal agreement as to how exactly to define the angle between two subspaces. There are diverse definitions each of which yields a different number(s).

- quickfur
- Pentonian
**Posts:**2984**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

I believe that for the intersection of two n dimensional objects in a 2N dimension space there will be n significant angles. Boiling that down to one angle may be of heuristic use in statistics but that's not what I want. Given one plane and the n significant angles it should uniquely identify the other plane.

Here is the question I have broadcast on the Internet.

Suppose I have two intersecting planes in a four dimensional Euclidean space. It seems to me that there are two angles between these planes. If the two planes intersect in a line then one of those angles is zero. If the two angles are non-zero then the planes intersect in a point. If one plane is the WX plane and the other the YZ plane then the two angles are both 90 degrees.

The best I can do is this. Each plane is defined by two orthonormal basis vectors. Project a basis vector onto the other plane. Then the angle is the arccos of the length of the projection. Do this with each basis vector to get the two angles. However I believe that the result depends on the choice of basis vectors.

What I need could be to find a unit vector in a plane so that it's projection onto the other plane is of maximum length. I don't know how to do that. I suppose find an expression for the length of the projection, take the derivative, and find where it is zero. Is there an easier way?

Here is the question I have broadcast on the Internet.

Suppose I have two intersecting planes in a four dimensional Euclidean space. It seems to me that there are two angles between these planes. If the two planes intersect in a line then one of those angles is zero. If the two angles are non-zero then the planes intersect in a point. If one plane is the WX plane and the other the YZ plane then the two angles are both 90 degrees.

The best I can do is this. Each plane is defined by two orthonormal basis vectors. Project a basis vector onto the other plane. Then the angle is the arccos of the length of the projection. Do this with each basis vector to get the two angles. However I believe that the result depends on the choice of basis vectors.

What I need could be to find a unit vector in a plane so that it's projection onto the other plane is of maximum length. I don't know how to do that. I suppose find an expression for the length of the projection, take the derivative, and find where it is zero. Is there an easier way?

- PatrickPowers
- Tetronian
**Posts:**478**Joined:**Wed Dec 02, 2015 1:36 am

Suppose this is going to be generalized to N dimensions. We have two hyperplanes, one with m dimensions and the other with n. To have the simplest cases suppose that m+n<=N. The first question is, how many significant angles are there?

Let's say we have a 2-plane and a 4-plane in 6D that intersect. Are there 2 significant angles or 4? My best guess is that there are four, but two of them always have the magnitude of pi/2 and hence of less interest.

Let's say we have a 2-plane and a 4-plane in 6D that intersect. Are there 2 significant angles or 4? My best guess is that there are four, but two of them always have the magnitude of pi/2 and hence of less interest.

- PatrickPowers
- Tetronian
**Posts:**478**Joined:**Wed Dec 02, 2015 1:36 am

You're right! There are two angles between two planes.

I haven't thought about this before; I only used one angle, given by the dot product of unit bivectors: cos α = |A•B|. This works in 3D, and it's at least not obviously wrong or undefined in higher dimensions.

As a unit vector v rotates in the plane B, its angle α from the plane A varies sinusoidally (rather, ||v•A||² = cos²α varies sinusoidally), with some minimum angle α₁ and maximum angle α₂.

The dot product of bivectors gives |A•B| = cos α₁ cos α₂, and the wedge product of bivectors gives ||A∧B|| = sin α₁ sin α₂. Taken together, these products determine α₁ and α₂ (assumed to be positive acute angles).

To make this more concrete, take A = e₁e₂ and B = (e₁ cos α₁ + e₃ sin α₁)(e₂ cos α₂ + e₄ sin α₂), where e₁,e₂,e₃,e₄ form an orthonormal basis for 4D space. Note that B is the geometric product of two orthonormal vectors f₁ and f₂:

f₁ = e₁ cos α₁ + e₃ sin α₁ = (cos α₁, 0 , sin α₁, 0 )

f₂ = e₂ cos α₂ + e₄ sin α₂ = ( 0 , cos α₂, 0 , sin α₂)

f₁•f₁ = cos²α₁ + sin²α₁ = 1, f₂•f₂ = cos²α₂ + sin²α₂ = 1, f₁•f₂ = 0

A = e₁e₂

B = f₁f₂ = e₁e₂ cos α₁ cos α₂ + e₁e₄ cos α₁ sin α₂ + e₃e₂ sin α₁ cos α₂ + e₃e₄ sin α₁ sin α₂

A•B = - cos α₁ cos α₂

A∧B = e₁e₂e₃e₄ sin α₁ sin α₂

Any unit vector in the plane B can be written as v = f₁ cos θ + f₂ sin θ, for some angle θ. Taking vector-bivector dot products:

v•A = (f₁•A) cos θ + (f₂•A) sin θ

= (e₂ cos α₁) cos θ + (-e₁ cos α₂) sin θ

= e₁ (-cos α₂ sin θ) + e₂ (cos α₁ cos θ) = (-cos α₂ sin θ, cos α₁ cos θ , 0 , 0 )

||v•A||² = cos²α₂ sin²θ + cos²α₁ cos²θ

= cos²α₂ (1 - cos 2θ)/2 + cos²α₁ (1 + cos 2θ)/2

= (cos²α₁ + cos²α₂)/2 + (cos²α₁ - cos²α₂)/2 cos 2θ

This varies sinusoidally with θ. Its maximum and minimum values are cos²α₁ and cos²α₂, achieved when θ is some multiple of 90°.

I haven't thought about this before; I only used one angle, given by the dot product of unit bivectors: cos α = |A•B|. This works in 3D, and it's at least not obviously wrong or undefined in higher dimensions.

As a unit vector v rotates in the plane B, its angle α from the plane A varies sinusoidally (rather, ||v•A||² = cos²α varies sinusoidally), with some minimum angle α₁ and maximum angle α₂.

The dot product of bivectors gives |A•B| = cos α₁ cos α₂, and the wedge product of bivectors gives ||A∧B|| = sin α₁ sin α₂. Taken together, these products determine α₁ and α₂ (assumed to be positive acute angles).

To make this more concrete, take A = e₁e₂ and B = (e₁ cos α₁ + e₃ sin α₁)(e₂ cos α₂ + e₄ sin α₂), where e₁,e₂,e₃,e₄ form an orthonormal basis for 4D space. Note that B is the geometric product of two orthonormal vectors f₁ and f₂:

f₁ = e₁ cos α₁ + e₃ sin α₁ = (cos α₁, 0 , sin α₁, 0 )

f₂ = e₂ cos α₂ + e₄ sin α₂ = ( 0 , cos α₂, 0 , sin α₂)

f₁•f₁ = cos²α₁ + sin²α₁ = 1, f₂•f₂ = cos²α₂ + sin²α₂ = 1, f₁•f₂ = 0

A = e₁e₂

B = f₁f₂ = e₁e₂ cos α₁ cos α₂ + e₁e₄ cos α₁ sin α₂ + e₃e₂ sin α₁ cos α₂ + e₃e₄ sin α₁ sin α₂

A•B = - cos α₁ cos α₂

A∧B = e₁e₂e₃e₄ sin α₁ sin α₂

Any unit vector in the plane B can be written as v = f₁ cos θ + f₂ sin θ, for some angle θ. Taking vector-bivector dot products:

v•A = (f₁•A) cos θ + (f₂•A) sin θ

= (e₂ cos α₁) cos θ + (-e₁ cos α₂) sin θ

= e₁ (-cos α₂ sin θ) + e₂ (cos α₁ cos θ) = (-cos α₂ sin θ, cos α₁ cos θ , 0 , 0 )

||v•A||² = cos²α₂ sin²θ + cos²α₁ cos²θ

= cos²α₂ (1 - cos 2θ)/2 + cos²α₁ (1 + cos 2θ)/2

= (cos²α₁ + cos²α₂)/2 + (cos²α₁ - cos²α₂)/2 cos 2θ

This varies sinusoidally with θ. Its maximum and minimum values are cos²α₁ and cos²α₂, achieved when θ is some multiple of 90°.

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**520**Joined:**Tue Sep 18, 2018 4:10 am

The dot product of bivectors gives |A•B| = cos α₁ cos α₂, and the wedge product of bivectors gives ||A∧B|| = sin α₁ sin α₂. Taken together, these products determine α₁ and α₂ (assumed to be positive acute angles).

Hmm, given cos α₁ cos α₂ and sin α₁ sin α₂ I don't see how to solve for α₁ and α₂. The magnitudes of the two expressions aren't independent so we essentially have one equation and two unknowns. On the other hand, this still is quite useful. The important boundary cases with sin α₁ sin α₂ = 0 or 1 are there. The case where we know that α₁ = α₂ is solved, as is the case when one of the angles is known. In case of two random planes about which we know nothing more, we can still get minimum values for the major and minor angles. In general sin α₁ sin α₂ gives one number that should give a good sense of the situation. If close to one then the planes are close to perpendicular. If the number is nearly zero then we know that the two planes are close to intersecting in a line. All and all I say that this answer is good enough for my purposes.

It's also quite useful to know that it varies sinusoidally. That means that the vectors associated with the angles are always perpendicular. Or at least, so I believe today.

By the way, it is possible for two planes to be both not parallel and not intersecting. They are parallel in only one dimension. If you have two random planes what is the probability they will intersect? I say the probability is one. But that is another story.

- PatrickPowers
- Tetronian
**Posts:**478**Joined:**Wed Dec 02, 2015 1:36 am

Here is something that I wrote explaining the two angle thing. I'm posting it here as part of the record.

I'm going to show that given two 2D planes in 4D we can define two angles between them. If both angles are small then the planes are close to being coplanar. If both angles are large then the planes are close to being perpendicular. If one is small and the other large then the planes are close together in one sense but not in the other. Given two proper angles we can construct two planes that have that relationship.

So what is my question? What I would like to do is the inverse of that. Given two planes, find the two angles.

First we will show that these two angles are always well-defined. Consider four free variables w,x,y,z. Then the sets [w,x,0,0] and [0,0,y,z] are planes that intersect only at the origin. Take any vector in one plane and project it into the other. You get a vector of length zero. We can say that the angle between the two planes is arccos(0)=pi/2. The two planes are perpendicular.

The sets [w,x,w,x] and [0,0,y,z] are also planes that intersect only at the origin. The projection of any vector in the first plane into the second is [0,0,w,x]. For any unit vector in the first plane we have 2(w^2+x^2) = 1. The length of the projection is 1/2. We have a well-defined angle of arccos(1/2) = pi/3.

The sets [w,x,w,0] and [0,0,y,z] are also planes that intersect only at the origin. The projection of any vector in the first plane into the second is [0,0,w,0]. For any unit vector in the first plane the length of the projection can be any number from 0 to 1/s, where s=sqrt(2). While not well-defined, the minimum and maximum of that length are well-defined. We get a major angle of arccos(0) = pi/2 and a minor of arccos(1/s) = pi/4.

Next have a constant a. The sets [w,x,aw,0] and [0,0,y,z] are also planes that intersect only at the origin. The projection of any vector in the first plane into the second is [0,0,aw,0]. We get a a major angle that is pi/2 and minor angle that depending on the value of a can be anything between 0 and pi/2.

Next have a constant b. Sets [w,x,aw,bx] and [0,0,y,z] are also planes that intersect only at the origin. The projection of any vector in the first plane into the second is [0,0,aw,bx]. Using the same process we get two angles. The point is that either angle could be anything from pi/2 to a limit of zero. We can make whatever we like, from perpendicular planes to a pair that is nearly coplanar but still intersect at a point.

We have shown that given two proper angles we can construct two planes that have that relationship.

I'm going to show that given two 2D planes in 4D we can define two angles between them. If both angles are small then the planes are close to being coplanar. If both angles are large then the planes are close to being perpendicular. If one is small and the other large then the planes are close together in one sense but not in the other. Given two proper angles we can construct two planes that have that relationship.

So what is my question? What I would like to do is the inverse of that. Given two planes, find the two angles.

First we will show that these two angles are always well-defined. Consider four free variables w,x,y,z. Then the sets [w,x,0,0] and [0,0,y,z] are planes that intersect only at the origin. Take any vector in one plane and project it into the other. You get a vector of length zero. We can say that the angle between the two planes is arccos(0)=pi/2. The two planes are perpendicular.

The sets [w,x,w,x] and [0,0,y,z] are also planes that intersect only at the origin. The projection of any vector in the first plane into the second is [0,0,w,x]. For any unit vector in the first plane we have 2(w^2+x^2) = 1. The length of the projection is 1/2. We have a well-defined angle of arccos(1/2) = pi/3.

The sets [w,x,w,0] and [0,0,y,z] are also planes that intersect only at the origin. The projection of any vector in the first plane into the second is [0,0,w,0]. For any unit vector in the first plane the length of the projection can be any number from 0 to 1/s, where s=sqrt(2). While not well-defined, the minimum and maximum of that length are well-defined. We get a major angle of arccos(0) = pi/2 and a minor of arccos(1/s) = pi/4.

Next have a constant a. The sets [w,x,aw,0] and [0,0,y,z] are also planes that intersect only at the origin. The projection of any vector in the first plane into the second is [0,0,aw,0]. We get a a major angle that is pi/2 and minor angle that depending on the value of a can be anything between 0 and pi/2.

Next have a constant b. Sets [w,x,aw,bx] and [0,0,y,z] are also planes that intersect only at the origin. The projection of any vector in the first plane into the second is [0,0,aw,bx]. Using the same process we get two angles. The point is that either angle could be anything from pi/2 to a limit of zero. We can make whatever we like, from perpendicular planes to a pair that is nearly coplanar but still intersect at a point.

We have shown that given two proper angles we can construct two planes that have that relationship.

- PatrickPowers
- Tetronian
**Posts:**478**Joined:**Wed Dec 02, 2015 1:36 am

PatrickPowers wrote:The dot product of bivectors gives |A•B| = cos α₁ cos α₂, and the wedge product of bivectors gives ||A∧B|| = sin α₁ sin α₂. Taken together, these products determine α₁ and α₂ (assumed to be positive acute angles).

Hmm, given cos α₁ cos α₂ and sin α₁ sin α₂ I don't see how to solve for α₁ and α₂. The magnitudes of the two expressions aren't independent so we essentially have one equation and two unknowns.

When I see a product of cosines and a product of sines, I think of the angle addition formula:

cos(α₁ + α₂) = cos α₁ cos α₂ - sin α₁ sin α₂ = |A•B| - ||A∧B||

cos(α₁ - α₂) = cos α₁ cos α₂ + sin α₁ sin α₂ = |A•B| + ||A∧B||

= cos(α₂ - α₁)

Since we're assuming that α₁ and α₂ are between 0° and 90° and that α₁ is the smaller one, both (α₁ + α₂) and (α₂ - α₁) are between 0° and 180°. Cosine is invertible there:

α₁ + α₂ = arccos(|A•B| - ||A∧B||)

α₂ - α₁ = arccos(|A•B| + ||A∧B||)

α₁ = (1/2) arccos(|A•B| - ||A∧B||) - (1/2) arccos(|A•B| + ||A∧B||)

α₂ = (1/2) arccos(|A•B| - ||A∧B||) + (1/2) arccos(|A•B| + ||A∧B||)

I suppose there are other ways to solve it, though.

PatrickPowers wrote:We have shown that given two proper angles we can construct two planes that have that relationship.

Yes, that's part of my previous post. Given any α₁ and α₂, we can construct one plane as the span of e₁=(1,0,0,0) and e₂=(0,1,0,0), and the other plane as the span of f₁=(cos α₁, 0 , sin α₁, 0 ) and f₂=( 0 , cos α₂, 0 , sin α₂).

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**520**Joined:**Tue Sep 18, 2018 4:10 am

(Can everyone see both of these types of angle brackets? Does either form look better? 〈1〉⟨2⟩ )

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**520**Joined:**Tue Sep 18, 2018 4:10 am

For general dimensions of A and B, consider the unit sphere in B. Its projection onto A is an ellipsoid. The cosines of the angles between A and B are just the radii of the ellipsoid.

Suppose A is a 5-dimensional subspace and B is a 3-dimensional subspace. There exists a pair of orthonormal bases for A and B, call them {a₁,a₂,a₃,a₄,a₅} and {b₁,b₂,b₃}, with these special properties: a₁•b₁ = cos α₁, a₂•b₂ = cos α₂, a₃•b₃ = cos α₃, and a•b = 0 otherwise (i.e. for non-matching indices).

Indeed, b₁ can be defined as the (or any) unit vector in B whose projection onto A has the maximum magnitude (this will be cos α₁). Then b₂ can be defined as the unit vector in the subspace of B orthogonal to b₁ whose projection onto A has the maximum magnitude (this will be cos α₂). Lastly, b₃ is either one of the two unit vectors in B orthogonal to both b₁ and b₂. It can be shown that the projections of b₁,b₂,b₃ onto A are still orthogonal to each other. Define a₁,a₂,a₃ as these projections, normalized to unit magnitude, assuming the projections are non-zero. The remaining vectors a₄,a₅, and any corresponding to projections being zero, can be chosen arbitrarily to complete the orthonormal basis for A.

This shows (in outline) that such a pair of bases exists, though it may not be easy to find, as several maximization problems must be solved.

Furthermore, if we write e₁=a₁, ... , e₅=a₅, and define e₆,e₇,e₈ as the (normalized) rejections of b₁,b₂,b₃ away from A, then we have an orthonormal basis {e₁, ... , e₈}, and the coordinates of our other vectors with respect to this basis are

a₁ = (1,0,0,0,0, 0,0,0),

a₂ = (0,1,0,0,0, 0,0,0),

a₃ = (0,0,1,0,0, 0,0,0),

a₄ = (0,0,0,1,0, 0,0,0),

a₅ = (0,0,0,0,1, 0,0,0),

b₁ = (cos α₁,0,0,0,0, sin α₁,0,0),

b₂ = (0,cos α₂,0,0,0, 0,sin α₂,0),

b₃ = (0,0,cos α₃,0,0, 0,0,sin α₃).

Now consider the grade 5 multivector A=a₁a₂a₃a₄a₅, and the grade 3 multivector B=b₁b₂b₃.

A = e₁e₂e₃e₄e₅

B = (e₁cos α₁ + e₆sin α₁) (e₂cos α₂ + e₇sin α₂) (e₃cos α₃ + e₈sin α₃)

= (e₁e₂e₃)(cos α₁ cos α₂ cos α₃) + (e₁e₂e₈)(cos α₁ cos α₂ sin α₃) + (e₁e₇e₃)(cos α₁ sin α₂ cos α₃) + (e₁e₇e₈)(cos α₁ sin α₂ sin α₃)

+ (e₆e₂e₃)(sin α₁ cos α₂ cos α₃) + (e₆e₂e₈)(sin α₁ cos α₂ sin α₃) + (e₆e₇e₃)(sin α₁ sin α₂ cos α₃) + (e₆e₇e₈)(sin α₁ sin α₂ sin α₃)

AB = (-e₄e₅)(cos α₁ cos α₂ cos α₃) + (-e₃e₄e₅e₈)(cos α₁ cos α₂ sin α₃) + (-e₂e₄e₅e₇)(cos α₁ sin α₂ cos α₃) + (e₂e₃e₄e₅e₇e₈)(cos α₁ sin α₂ sin α₃)

+ (-e₁e₄e₅e₆)(sin α₁ cos α₂ cos α₃) + (e₁e₃e₄e₅e₆e₈)(sin α₁ cos α₂ sin α₃) + (e₁e₂e₄e₅e₆e₇)(sin α₁ sin α₂ cos α₃) + (e₁e₂e₃e₄e₅e₆e₇e₈)(sin α₁ sin α₂ sin α₃)

⟨AB⟩₂ = (-e₄e₅)(cos α₁ cos α₂ cos α₃)

⟨AB⟩₄ = (-e₃e₄e₅e₈)(cos α₁ cos α₂ sin α₃) + (-e₂e₄e₅e₇)(cos α₁ sin α₂ cos α₃) + (-e₁e₄e₅e₆)(sin α₁ cos α₂ cos α₃)

⟨AB⟩₆ = (e₂e₃e₄e₅e₇e₈)(cos α₁ sin α₂ sin α₃) + (e₁e₃e₄e₅e₆e₈)(sin α₁ cos α₂ sin α₃) + (e₁e₂e₄e₅e₆e₇)(sin α₁ sin α₂ cos α₃)

⟨AB⟩₈ = (e₁e₂e₃e₄e₅e₆e₇e₈)(sin α₁ sin α₂ sin α₃)

||⟨AB⟩₂||² = cos²α₁ cos²α₂ cos²α₃

||⟨AB⟩₄||² = cos²α₁ cos²α₂ sin²α₃ + cos²α₁ sin²α₂ cos²α₃ + sin²α₁ cos²α₂ cos²α₃

||⟨AB⟩₆||² = cos²α₁ sin²α₂ sin²α₃ + sin²α₁ cos²α₂ sin²α₃ + sin²α₁ sin²α₂ cos²α₃

||⟨AB⟩₈||² = sin²α₁ sin²α₂ sin²α₃

(The point of using these multivector magnitudes is that they can be calculated in any basis, without needing to know those special a and b vectors.)

This system doesn't look easy to solve for α₁,α₂,α₃. However, there are some simplifications to be made.

The intersection of A and B, and the subspace of B which is orthogonal to all of A, and the subspace of A which is orthogonal to all of B, can be calculated using basic linear algebra. The intersection has all the 0° angles, and the other two have all the 90° angles. Orthogonal to those three subspaces, the remaining subspaces of A and B have all the angles strictly between 0° and 90°. If the remaining subspaces are 0D or 1D or 2D, we already know how to find the angles between them. So assume they're 3D. This implies that α₁,α₂,α₃ are strictly between 0° and 90°, so the sines and cosines are all non-zero, and we can divide:

||⟨AB⟩₄||²/||⟨AB⟩₂||² = tan²α₃ + tan²α₂ + tan²α₁

||⟨AB⟩₆||²/||⟨AB⟩₂||² = tan²α₂ tan²α₃ + tan²α₁ tan²α₃ + tan²α₁ tan²α₂

||⟨AB⟩₈||²/||⟨AB⟩₂||² = tan²α₁ tan²α₂ tan²α₃

I recognize these as symmetric polynomials. By Vieta's formulas, if t is any of these squared tangents then

t³ - t² ||⟨AB⟩₄||²/||⟨AB⟩₂||² + t ||⟨AB⟩₆||²/||⟨AB⟩₂||² - ||⟨AB⟩₈||²/||⟨AB⟩₂||² = 0.

This is a nice single-variable algebraic equation that we can try to solve for t. Then the angles are α=arctan√t.

Of course all of this generalizes. The dimensions 5 and 3 were chosen to show the general pattern without being too abstract or unwieldy.

Suppose A is a 5-dimensional subspace and B is a 3-dimensional subspace. There exists a pair of orthonormal bases for A and B, call them {a₁,a₂,a₃,a₄,a₅} and {b₁,b₂,b₃}, with these special properties: a₁•b₁ = cos α₁, a₂•b₂ = cos α₂, a₃•b₃ = cos α₃, and a•b = 0 otherwise (i.e. for non-matching indices).

Indeed, b₁ can be defined as the (or any) unit vector in B whose projection onto A has the maximum magnitude (this will be cos α₁). Then b₂ can be defined as the unit vector in the subspace of B orthogonal to b₁ whose projection onto A has the maximum magnitude (this will be cos α₂). Lastly, b₃ is either one of the two unit vectors in B orthogonal to both b₁ and b₂. It can be shown that the projections of b₁,b₂,b₃ onto A are still orthogonal to each other. Define a₁,a₂,a₃ as these projections, normalized to unit magnitude, assuming the projections are non-zero. The remaining vectors a₄,a₅, and any corresponding to projections being zero, can be chosen arbitrarily to complete the orthonormal basis for A.

This shows (in outline) that such a pair of bases exists, though it may not be easy to find, as several maximization problems must be solved.

Furthermore, if we write e₁=a₁, ... , e₅=a₅, and define e₆,e₇,e₈ as the (normalized) rejections of b₁,b₂,b₃ away from A, then we have an orthonormal basis {e₁, ... , e₈}, and the coordinates of our other vectors with respect to this basis are

a₁ = (1,0,0,0,0, 0,0,0),

a₂ = (0,1,0,0,0, 0,0,0),

a₃ = (0,0,1,0,0, 0,0,0),

a₄ = (0,0,0,1,0, 0,0,0),

a₅ = (0,0,0,0,1, 0,0,0),

b₁ = (cos α₁,0,0,0,0, sin α₁,0,0),

b₂ = (0,cos α₂,0,0,0, 0,sin α₂,0),

b₃ = (0,0,cos α₃,0,0, 0,0,sin α₃).

Now consider the grade 5 multivector A=a₁a₂a₃a₄a₅, and the grade 3 multivector B=b₁b₂b₃.

A = e₁e₂e₃e₄e₅

B = (e₁cos α₁ + e₆sin α₁) (e₂cos α₂ + e₇sin α₂) (e₃cos α₃ + e₈sin α₃)

= (e₁e₂e₃)(cos α₁ cos α₂ cos α₃) + (e₁e₂e₈)(cos α₁ cos α₂ sin α₃) + (e₁e₇e₃)(cos α₁ sin α₂ cos α₃) + (e₁e₇e₈)(cos α₁ sin α₂ sin α₃)

+ (e₆e₂e₃)(sin α₁ cos α₂ cos α₃) + (e₆e₂e₈)(sin α₁ cos α₂ sin α₃) + (e₆e₇e₃)(sin α₁ sin α₂ cos α₃) + (e₆e₇e₈)(sin α₁ sin α₂ sin α₃)

AB = (-e₄e₅)(cos α₁ cos α₂ cos α₃) + (-e₃e₄e₅e₈)(cos α₁ cos α₂ sin α₃) + (-e₂e₄e₅e₇)(cos α₁ sin α₂ cos α₃) + (e₂e₃e₄e₅e₇e₈)(cos α₁ sin α₂ sin α₃)

+ (-e₁e₄e₅e₆)(sin α₁ cos α₂ cos α₃) + (e₁e₃e₄e₅e₆e₈)(sin α₁ cos α₂ sin α₃) + (e₁e₂e₄e₅e₆e₇)(sin α₁ sin α₂ cos α₃) + (e₁e₂e₃e₄e₅e₆e₇e₈)(sin α₁ sin α₂ sin α₃)

⟨AB⟩₂ = (-e₄e₅)(cos α₁ cos α₂ cos α₃)

⟨AB⟩₄ = (-e₃e₄e₅e₈)(cos α₁ cos α₂ sin α₃) + (-e₂e₄e₅e₇)(cos α₁ sin α₂ cos α₃) + (-e₁e₄e₅e₆)(sin α₁ cos α₂ cos α₃)

⟨AB⟩₆ = (e₂e₃e₄e₅e₇e₈)(cos α₁ sin α₂ sin α₃) + (e₁e₃e₄e₅e₆e₈)(sin α₁ cos α₂ sin α₃) + (e₁e₂e₄e₅e₆e₇)(sin α₁ sin α₂ cos α₃)

⟨AB⟩₈ = (e₁e₂e₃e₄e₅e₆e₇e₈)(sin α₁ sin α₂ sin α₃)

||⟨AB⟩₂||² = cos²α₁ cos²α₂ cos²α₃

||⟨AB⟩₄||² = cos²α₁ cos²α₂ sin²α₃ + cos²α₁ sin²α₂ cos²α₃ + sin²α₁ cos²α₂ cos²α₃

||⟨AB⟩₆||² = cos²α₁ sin²α₂ sin²α₃ + sin²α₁ cos²α₂ sin²α₃ + sin²α₁ sin²α₂ cos²α₃

||⟨AB⟩₈||² = sin²α₁ sin²α₂ sin²α₃

(The point of using these multivector magnitudes is that they can be calculated in any basis, without needing to know those special a and b vectors.)

This system doesn't look easy to solve for α₁,α₂,α₃. However, there are some simplifications to be made.

The intersection of A and B, and the subspace of B which is orthogonal to all of A, and the subspace of A which is orthogonal to all of B, can be calculated using basic linear algebra. The intersection has all the 0° angles, and the other two have all the 90° angles. Orthogonal to those three subspaces, the remaining subspaces of A and B have all the angles strictly between 0° and 90°. If the remaining subspaces are 0D or 1D or 2D, we already know how to find the angles between them. So assume they're 3D. This implies that α₁,α₂,α₃ are strictly between 0° and 90°, so the sines and cosines are all non-zero, and we can divide:

||⟨AB⟩₄||²/||⟨AB⟩₂||² = tan²α₃ + tan²α₂ + tan²α₁

||⟨AB⟩₆||²/||⟨AB⟩₂||² = tan²α₂ tan²α₃ + tan²α₁ tan²α₃ + tan²α₁ tan²α₂

||⟨AB⟩₈||²/||⟨AB⟩₂||² = tan²α₁ tan²α₂ tan²α₃

I recognize these as symmetric polynomials. By Vieta's formulas, if t is any of these squared tangents then

t³ - t² ||⟨AB⟩₄||²/||⟨AB⟩₂||² + t ||⟨AB⟩₆||²/||⟨AB⟩₂||² - ||⟨AB⟩₈||²/||⟨AB⟩₂||² = 0.

This is a nice single-variable algebraic equation that we can try to solve for t. Then the angles are α=arctan√t.

Of course all of this generalizes. The dimensions 5 and 3 were chosen to show the general pattern without being too abstract or unwieldy.

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**520**Joined:**Tue Sep 18, 2018 4:10 am

PatrickPowers wrote:Let's say we have a 2-plane and a 4-plane in 6D that intersect. Are there 2 significant angles or 4? My best guess is that there are four, but two of them always have the magnitude of pi/2 and hence of less interest.

Correct.

If the unit sphere in a larger subspace is projected onto a smaller subspace, then some of the ellipsoid's radii are 0, and the corresponding angles are 90°. If the unit sphere in the smaller subspace is projected onto the larger subspace, then the ellipsoid may or may not have radii being 0. The non-zero radii of the two ellipsoids are the same. In other words, the non-right angles between A and B are the same as the non-right angles between B and A.

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**520**Joined:**Tue Sep 18, 2018 4:10 am

Well that's amazing. I don't know what more to say.

- PatrickPowers
- Tetronian
**Posts:**478**Joined:**Wed Dec 02, 2015 1:36 am

mr_e_man wrote:For general dimensions of A and B, consider the unit sphere in B. Its projection onto A is an ellipsoid. The cosines of the angles between A and B are just the radii of the ellipsoid.

In other words, the cosines are the singular values of projecting from B to A.

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**520**Joined:**Tue Sep 18, 2018 4:10 am

Related posts from a year ago: viewtopic.php?f=11&t=2555

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**520**Joined:**Tue Sep 18, 2018 4:10 am

And here's another page I just found: https://en.wikipedia.org/wiki/Angles_between_flats

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**520**Joined:**Tue Sep 18, 2018 4:10 am

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